[Haskell-cafe] Should "do 1" compile

[Neil Mitchell]

Hi

> > foo = do (1 :: Int)
>
> While intuitively this should be disallowed, it seems a pity that
> desugaring couldn't be totally separated from typechecking. Hmm.

You can always desugar as:

do x ==> return () >> x

Although then you are relying on the Monad laws more than you possibly
should. You could also have:

monady :: Monad m => m a -> m a
monady x = x

do x ==> monady x

Then you are relying on an additional function that doesn't currently exist.

Thanks

Neil